Your Swine manure pit is rectangular measuring 50' by 100' and is 4' deep. If you use a 1" diameter hose 2 hours each day at 8 ft per second velocity to remove the manure from your swine pins how many days worth of storage space do you have in your pit? In other words how often will you have to pump out your pit? Assuming no loss to evaporation?
The amount of material is volume rate times time.
Volume= 8ft/sec*areahose*time
solve for time.
areahose is the cross sectional area of the one inch diameter hose.
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To find the time it takes to pump out the manure pit, we need to calculate the volume of manure being removed per day and then divide it by the volume of the pit.
First, let's calculate the area of the hose:
The diameter of the hose is 1", so the radius is 1/2 inch, which is 0.04167 feet (Since 1 foot = 12 inches).
The cross-sectional area of the hose is πr^2, where r is the radius:
Area_hose = π * (0.04167)^2 = 0.005454 ft^2
Next, let's calculate the volume of manure being removed per second:
The velocity of the manure flowing through the hose is 8 ft/sec.
Volume_rate = velocity * area_hose = 8 ft/sec * 0.005454 ft^2 = 0.043632 ft^3/sec
Now, let's find the volume of manure being removed per hour:
There are 60 seconds in a minute and 60 minutes in an hour.
Volume_hour = volume_rate * (60 sec/min * 60 min/hour) = 0.043632 ft^3/sec * 3600 sec/hour = 157.0752 ft^3/hour
And finally, let's calculate the volume of the manure pit:
The pit is rectangular measuring 50' by 100' and is 4' deep.
Volume_pit = length * width * depth = 50 ft * 100 ft * 4 ft = 20000 ft^3
Now we can find the number of days worth of storage space in the pit:
Days_of_storage = Volume_pit / Volume_hour
Days_of_storage = 20000 ft^3 / 157.0752 ft^3/hour
Days_of_storage ≈ 127.27 hours / 24 hours/day ≈ 5.30 days
Therefore, you will need to pump out your pit approximately every 5.30 days to maintain the same level assuming no loss to evaporation.